zoukankan      html  css  js  c++  java
  • POJ-3255-Roadblocks(次短路的另一种求法)

    Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.

    The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.

    The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).

    Input

    Line 1: Two space-separated integers: N and R 
    Lines 2.. R+1: Each line contains three space-separated integers: AB, and D that describe a road that connects intersections A and B and has length D (1 ≤ D ≤ 5000)

    Output

    Line 1: The length of the second shortest path between node 1 and node N

    Sample Input

    4 4
    1 2 100
    2 4 200
    2 3 250
    3 4 100

    Sample Output

    450

    Hint

    Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    #include<queue>
    #include<stack>
    #include<set>
    #include<map>
    #include<vector>
    #include<cmath>
    
    
    const int maxn=1e5+5;
    typedef long long ll;
    using namespace std;
    
    struct node
    {
       int pos,w;
       node(int x,int y)
       {
           pos=x;
           w=y;
       }
       bool friend operator<(node x,node y )
       {
           return x.w>y.w;
       }
    };
    struct edge
    {
       int u,v;
       ll cost;
       int nxt;
    }Edge[maxn<<1];
    int cnt;
    
    ll dis[5005],dis2[5005];
    int head[5005],vis[5005];
    void Add(int u,int v,ll w)
    {
       Edge[cnt].u=u;
       Edge[cnt].v=v;
       Edge[cnt].cost=w;
       Edge[cnt].nxt=head[u];
       head[u]=cnt++;
    }
    void Dijkstra(int u)
    {
        dis[u]=0;
        priority_queue<node>q;
        q.push(node(u,0));
        while(!q.empty())
        {
           node now=q.top();
           q.pop();
           if(vis[now.pos])
           {
               continue;
           }
           vis[now.pos]=1;
           for(int t=head[now.pos];t!=-1;t=Edge[t].nxt)
           {
               if(dis[now.pos]+Edge[t].cost<dis[Edge[t].v])
               {
                   dis[Edge[t].v]=dis[now.pos]+Edge[t].cost;
                   q.push(node(Edge[t].v,dis[Edge[t].v]));
               }
           }
        }
        return ;
    }
    int main()
    {
        int n,m;
        scanf("%d%d",&n,&m);
        memset(vis,0,sizeof(vis));
        memset(head,-1,sizeof(head));
        memset(dis,0x3f3f3f3f,sizeof(dis));
        cnt=0;
        int u,v;
        ll w;
        for(int t=0;t<m;t++)
        {
            scanf("%d%d%lld",&u,&v,&w);
            Add(u,v,w);
            Add(v,u,w);
        }
        Dijkstra(1);
        for(int t=1;t<=n;t++)
        {
            dis2[t]=dis[t];
        }
        memset(dis,0x3f3f3f3f,sizeof(dis));
        memset(vis,0,sizeof(vis));
        Dijkstra(n);
        int ans=0x3f3f3f3f;
        for(int t=0;t<cnt;t++)
        {
            if(dis[Edge[t].u]+dis2[Edge[t].v]+Edge[t].cost<ans&&(dis[Edge[t].u]+dis2[Edge[t].v]+Edge[t].cost)!=dis2[n])
            {
                ans=dis[Edge[t].u]+dis2[Edge[t].v]+Edge[t].cost;
            }
        }
        printf("%d
    ",ans);
        system("pause");
        return 0;
    }
    
    
  • 相关阅读:
    easyui属性表格的一点小总结
    我是程序员:国庆带女朋友回家见父母
    我睁着朦胧呆滞的双眼立在初秋的夜里
    iframe父子兄弟之间调用传值(contentWindow && parent)
    向前兼容和向后兼容的含义
    char 与 String 相等比较
    2012国庆12天的长假
    JavaScript遍历XML总结
    Spring3.x企业应用开发实战Spring+Hibernat架构分析
    w3school上系统过了一遍Jquery的总结
  • 原文地址:https://www.cnblogs.com/Staceyacm/p/11337315.html
Copyright © 2011-2022 走看看